Problem 20 - Advanced Engineering Mathematics

Question

Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(\frac{1}{t}+\frac{1}{t^{2}}-\frac{y}{t^{2}+y^{2}}\right) d t+\left(y e^{y}+\frac{1}{t^{2}+y^{2}}\right) d y=0 $$

Step-by-Step Solution

Answer

The differential equation is not exact.

Step 1: Identifying Functions P and Q

In the given differential equation, \[ M(t, y) = \left(\frac{1}{t}+\frac{1}{t^{2}}-\frac{y}{t^{2}+y^{2}}\right) \] is the coefficient of \( dt \), and \[ N(t, y) = \left(y e^{y}+\frac{1}{t^{2}+y^{2}}\right) \] is the coefficient of \( dy \).

Step 2: Checking for Exactness

To check if the equation is exact, we need to verify if \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}. \]First, calculate \[ \frac{\partial M}{\partial y} = \frac{0 - (t^2 + y^2)(1)t^2 + y^2)}{(t^2 + y^2)^2} = \frac{t^2}{(t^2 + y^2)^2}.\]Next, calculate \[ \frac{\partial N}{\partial t} = -\frac{2t}{(t^2 + y^2)^2}.\]Since the partial derivatives are not equal, the equation is not exact.

Step 3: Conclusion

The given differential equation is not exact since \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial t} \). Therefore, we cannot solve it using methods for exact equations.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of an unknown function. These types of equations are vital because they describe various physical phenomena, such as motion, heat, or waves. The given equation \[\left(\frac{1}{t}+\frac{1}{t^{2}}-\frac{y}{t^{2}+y^{2}}\right) dt+\left(y e^{y}+\frac{1}{t^{2}+y^{2}}\right) dy=0\]is a special type known as a first-order differential equation because it only includes first derivatives. To solve such an equation, we need to analyze its structure: identifying functions and determining conditions like exactness. This involves breaking down the equation into parts and testing for certain mathematical properties, such as exactness, to explore possible solution paths.

Partial Derivatives

Partial derivatives play a crucial role when dealing with differential equations that involve multiple variables, like our example function \(M(t, y)\) and \(N(t, y)\). A partial derivative measures how a function changes when one of several variables changes, keeping others constant.
For the given equation, we need to compute \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\) to check the equation's exactness. This involves differentiating the function with respect to various variables while treating other variables as constant.

• For \(\frac{\partial M}{\partial y}\), we calculate the derivative of \(M\) with respect to \(y\).
• Similarly, \(\frac{\partial N}{\partial t}\) is the derivative of \(N\) with respect to \(t\).

These calculations help determine whether the differential equation meets the exactness condition. If they are not equal, as in this example, it indicates the differential equation is not exact, meaning it does not have straightforward solutions using methods for exact equations.

Exactness Condition

The exactness condition is a specific criterion used to test whether a differential equation can be solved straightforwardly. It involves verifying that the mixed partial derivatives of certain functions are equal. In the given equation, the functions \(M(t, y)\) and \(N(t, y)\) need to satisfy this condition: \[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\]This means the changes in the function in one direction should match the changes in the perpendicular direction, intuitively suggesting a balance or symmetry. In practical terms, if the condition holds, the equation is termed "exact." A bit like ensuring different routes lead to the same destination.
However, since these derivatives are not equal in the example provided \((\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial t})\), the equation is not exact. Therefore, it can't be solved by standard methods that work on exact equations. In cases like these, alternative strategies, such as finding an integrating factor or using numerical methods, may be necessary to explore solutions.